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## Hot answers tagged diophantine-approximation

318

The following is exact. :-) $$\phi=\frac{\frac{\pi}{\pi}+\sqrt{\frac{e+e+e+e+e}{e}}}{\frac{e}{e}+\frac{\pi}{\pi}}$$

197

$e$ and $\pi$ are transcendental numbers, that is to say they are not the solution of any polynomial with rational coefficients. It's not hard to see that if $x$ is transcendental, then the following are also transcendental: $x \pm c$ for any rational number $c$, $kx$ for any nonzero rational number $k$ (so $x/k$ too), $x^n$ for any whole number $n > ... 148 At the time of writing, three of the other answers simply express the golden ratio by using expressions like$e/e$and$\pi/\pi$to get small integers. The fourth and final one discusses why a good solution is unlikely. I believe using the imaginary unit$i=\sqrt{-1}$results in the following very elegant solution: $$\varphi = e^{i\pi/5} + e^{-i\pi/5}.$$ ... 71 If the discussion is not limited to closed-form expressions, it's worth adding that Ramanujan's first letter to Hardy contains an identity that, with a slight rearrangement, allows one to precisely express$\phi$in terms of$\pi$and$e$:$\phi =\sqrt{\frac{1}{2} \left(5+\sqrt{5}\right)}-\cfrac{e^{-\frac{2 \pi}{5}}}{1+\cfrac{e^{-2\pi}} {1+\cfrac{e^{-4\pi}} ...

32

Hint: $|\sqrt2 -1|<1/2$, so as $n\to\infty$ we have that $(\sqrt2-1)^n\to ?$ In addition to that use the fact that the set $S$ is a ring, i.e. closed under multiplication and addition.

29

Rather than just give you a fish, I'll teach you how to fish: $(\phi - 1)\phi = 1$ $\phi^2 - \phi - 1 = 0$ $\phi = \dfrac{1 + \sqrt{5}}{2}$ Now replace the integers there with a load of self-cancelling $\pi$/$e$ terms which ultimately give you the values 1, 5, (2 or 4) to taste (-: Throw in some complex numbers too if you're feeling brave For example: ...

26

This is problem 11162, posed by Paolo Perfetti, in the June-July 2005 issue of the American Mathematical Monthly. The solution below, due to the Microsoft Research Problems Group, is found in the February 2007 issue of the same magazine. For positive integer $n$, define $$A_n=[0,2^n)\cap\{k\in \mathbb{N}:|\!\sin k|<\textstyle{1\over n}\},\quad ... 24 \sqrt e \approx 1.64872 is not "very close" to \phi \approx 1.61803. Here is a very good approximation:$$ \phi \approx \frac{1967981\,\pi-314270\,e}{3293083} $$The error is about 2 \times 10^{-16}. This relation was found using FindIntegerNullVector[N@{Pi,E,(1+Sqrt[5])/2}] with Mathematica (sadly, Wolfram Alpha does not understand this). 22 Supose not, so that there exists an \varepsilon>0 such that (0,\varepsilon)\cap S=\emptyset. \qquad\qquad\qquad(\star) It follows that \alpha=\inf S\cap(0,+\infty) is a positive number. The choice of \alpha and its positivity implies that the one and only element of S which is in [0,\alpha) is 0. I claim that \alpha\in S. ... 19 Hint: \pi is not a Liouville number, so there exists m\in\mathbb{N} such that for all p,q\in\mathbb{Z} with q>1, we have$$ \left| \pi - \frac{p}{q}\right| \geq \frac{1}{q^m}.$$This should allow you to keep \sin n away from 0. Edit: Full Solution: Let m be as above. So for all p,q\in\mathbb{Z} with q>1 we have$$ \left| \pi - ...

19

RIES can find solutions to approximation problems like this. Running ries -NlLeE -s -l6 2.718281828459045235360287471 (which says "find $e$ without using logarithms, exponentials, or e itself" -- $\pi$ and $\phi$ are already in by default) gives, among others, $$\phi\approx e-\sqrt[3]{\pi}$$ $$\phi\approx\frac{5(1+\pi)}{e}-6 \qquad \text{(to 7 decimal ... 18 Consider x = \sum_{j=1}^\infty a_j/4^j where each a_j is either 1 or 2. Thus the binary expansion of x consists of two-digit blocks which are either 10 or 01. Then x is far. But there are uncountably many choices, so all but countably many of them are irrational. 17 This isn't exceptionally good compared to the partial convergents of the continued fraction expansion. Terminating the continued fraction for \pi right before the 292 gives \frac{355}{113}= \textbf{3.141592}9203\ldots, which gets 6 digits after the decimal place right, while your fraction only gets 3 digits after the decimal place correct even ... 16 claim. The series \sum\limits_{n=1}^\infty \sin^n(n) diverges. Lemma. For all number x irrational there exist a rational sequence \{\frac{p_n}{q_n}\} where \{q_n\} is odd such that$$ \left\vert x-\frac{p_n}{q_n}\right\vert<\frac{1}{q_n^2} $$Proof. Define x_n=\frac{1}{x_{n-1}-\lfloor x_{n-1}\rfloor}. Let a_0=\lfloor x\rfloor and ... 15 An approximation:$$\phi \approx \frac { 7\pi }{ 5e } =1.618018$$12 If you define the sequence a_1 = a_2 = -e^{i\pi}, a_k = a_{k-1} + a_{k-2}, then \lim_{n \rightarrow \infty} \frac {a_{n+1}}{a_n} = \phi. 12 Here is another suggestion:$$\phi=\frac{\pi}{\pi+\pi}+\sqrt{\frac{e+e+e+e+e}{e+e+e+e}}$$12 Using the property stated in that article:$$\mu(x)=2 + \limsup \frac{\log a_{n+1}}{\log q_n}$$where the continued fraction expansion for x is [a_0,a_1,...] and the nth convergent is \frac{p_n}{q_n}. Start with a_0=2 and a_1=2, so q_0=1, q_1=2. Now, assume you have a continued fraction$$\frac{p_n}{q_n}=[a_0,...,a_n]$$Define a_{n+1} ... 12 Actually, we can do it for higher powers as well. Let F_k (N) denote the sum of the k^{th} powers of the differences in the Farey sequence. Then it is easy to see that$$F_0(N)=\sum_{n=1}^N \phi(n)\sim \frac{3N^2}{\pi^2}$$and$$F_1(N)=1.$$It seems you are looking for F_2 (N), and, you are correct that the asymptotic is \frac{\log N}{N^2}. More ... 12 The following shows that for n=3 the ratio can be as close to 1 as we like. For any positive integer m let: a = 3m^3 + 3m^2 + 2m + 1 b = 3m^3 + 3m^2 + 2m c = 3m^2 + 2m + 1 d = m Then it is readily shown that a^3 = b^3 + c^3 + d^3. If we focus on the 'wrong solution' a^3 = b^3 + c^3, the ratio of LHS to RHS is: ... 11 It converges to 0, in fact \frac{1}{k^7\sin{k}} already converges to 0, see Theorem 2 here. This theorem gives a nice characterization of the irrationality measure of \pi as the borderline number \mu such that \frac{1}{k^{u-1}\sin{k}} converges to 0 for u>\mu, and diverges for u<\mu. So \frac{1}{k^7\sin{k}} converges because \mu ... 11 Yes, the number e=\exp(1) is such a number. Also, maybe more famously \exp(-1). 11 It seems all the answers so far approaching this from a theoretical perspective are approaching this in terms of exact answers, but we can say a lot about when good approximations are possible too. Of course, some answers have already provided silly ways to do this exactly, so approximations may seem unnecessary, but it provides a nice avenue for some basic ... 10 There are some details that I haven't fully vetted but here's a long sketch which I believe should show divergence. Define c_n := \frac{n}{2\pi} \pmod 1 and let [a,b] be any interval in the torus \mathbb R/\mathbb Z. The discrepancy D(N) of the sequence c_n is defined to be the difference between \# \{ n \le N : c_n \in [a,b]\} and the expected ... 10 Though it is clear what would count as a positive answer, the question is not precise enough that to make clear what counts as a negative answer. Nonetheless, I believe that no similar formula could possibly be equivalent to denseness. The problem is that there are sequences that are dense but that are wildly non-equidistributed. For example, your formula ... 9 Let 0 < \epsilon \ll 1 be any small and M \gg 1 be any large positive numbers. Let \lambda be a number of the form a + b\sqrt{2} + c\sqrt{3} where a, b, c \in \mathbb{Z}, not all zero such that$$|a|, |b|, |c| < M \quad\text{ and }\quad |\lambda| = |a + b\sqrt{2}+c\sqrt{3}| < \epsilon $$It is easy to see \lambda is a root of the ... 8 Taking a step into generalization, it is true that every additive subgroup G of \mathbb R is either discrete or dense. This can be proved by considering \alpha = \inf \{ x \in G : x>0 \}. Then G is discrete iff \alpha >0, in which case G=\alpha \mathbb Z. In your case, Jyrki's suggestion implies that \alpha=0 and so S is dense. 8 When I first came across your question, I thought it was a modern-day approximation by somebody using a computer. But when d125q pointed out it was by Ramanujan, then I figured out he must have used a systematic method. One way is to use a Ramanujan-Sato pi formula like,$$\frac{1}{\pi} = \frac{1}{16}\sum_{n=0}^\infty ...

8

It is an overkill for sure, but since the continued fraction of $\sqrt{3}$ is given by: $$\sqrt{3}=[1;\overline{1,2}],\tag{1}$$ we have that: $$[1,1,2,1,2,1,2] = \frac{71}{41} \tag{2}$$ is an accurate approximation, and: $$\left|\sqrt{3}-\frac{71}{41}\right|\leq\frac{1}{41^2}<\frac{1}{1000}.\tag{3}$$

8

normal $\; \not\Rightarrow \;$ irrationality measure $2$ There exist numbers that are normal with irrationality measure $>2$. In fact, there exist normal numbers (meaning normal with respect to every base) that have irrationality measure $\infty.$ This is Theorem 2 in Bugeaud [2] (2002). For related results, see [1] and [3]. irrationality measure $2$ ...

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